The masses of three copper wires are in the ratio 2:3:5 and their lengths are in the ratio 5:3:2. Then, the ratio of their electrical resistance is

Options

A. `1:9:15`
B. `2:3:5`
C. `5:3:2`
D. `125:30:8`

Answer

To find the ratio of the electrical resistance of the three copper wires, we can use the formula for resistance:

\[ R = \frac{\rho L}{A} \]

Where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity (which is constant for copper),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 1: Understand the given ratios
The masses of the three wires are in the ratio \( 2:3:5 \) and their lengths are in the ratio \( 5:3:2 \).

### Step 2: Express the lengths
Let the lengths of the wires be:
- \( L_1 = 5x \)
- \( L_2 = 3x \)
- \( L_3 = 2x \)

### Step 3: Express the masses
Let the masses of the wires be:
- \( m_1 = 2y \)
- \( m_2 = 3y \)
- \( m_3 = 5y \)

### Step 4: Relate mass to volume and area
The mass of the wire can be expressed in terms of volume and density:
\[ m = \rho_{density} \cdot V \]
Where \( V = A \cdot L \). Thus:
\[ m = \rho_{density} \cdot A \cdot L \]

From this, we can express the area \( A \) in terms of mass and length:
\[ A = \frac{m}{\rho_{density} \cdot L} \]

### Step 5: Calculate the areas for each wire
Substituting the expressions for mass and length into the area formula:

1. For wire 1:
\[ A_1 = \frac{2y}{\rho_{density} \cdot 5x} = \frac{2y}{5\rho_{density}x} \]

2. For wire 2:
\[ A_2 = \frac{3y}{\rho_{density} \cdot 3x} = \frac{3y}{3\rho_{density}x} = \frac{y}{\rho_{density}x} \]

3. For wire 3:
\[ A_3 = \frac{5y}{\rho_{density} \cdot 2x} = \frac{5y}{2\rho_{density}x} \]

### Step 6: Substitute into the resistance formula
Now, we can substitute the lengths and areas into the resistance formula:

1. For wire 1:
\[ R_1 = \frac{\rho \cdot 5x}{A_1} = \frac{\rho \cdot 5x}{\frac{2y}{5\rho_{density}x}} = \frac{25\rho \cdot x^2 \cdot \rho_{density}}{2y} \]

2. For wire 2:
\[ R_2 = \frac{\rho \cdot 3x}{A_2} = \frac{\rho \cdot 3x}{\frac{y}{\rho_{density}x}} = \frac{3\rho \cdot x^2 \cdot \rho_{density}}{y} \]

3. For wire 3:
\[ R_3 = \frac{\rho \cdot 2x}{A_3} = \frac{\rho \cdot 2x}{\frac{5y}{2\rho_{density}x}} = \frac{4\rho \cdot x^2 \cdot \rho_{density}}{5y} \]

### Step 7: Find the ratio of the resistances
Now we can find the ratio \( R_1 : R_2 : R_3 \):

\[ R_1 : R_2 : R_3 = \frac{25\rho \cdot x^2 \cdot \rho_{density}}{2y} : \frac{3\rho \cdot x^2 \cdot \rho_{density}}{y} : \frac{4\rho \cdot x^2 \cdot \rho_{density}}{5y} \]

Cancelling out common terms \( \rho, x^2, \rho_{density}, y \):

\[ R_1 : R_2 : R_3 = \frac{25}{2} : 3 : \frac{4}{5} \]

### Step 8: Convert to a common denominator
To simplify, we can convert these ratios to a common denominator:

- The common denominator for \( 2, 3, \frac{4}{5} \) is \( 30 \).
- Thus:
- \( R_1 = \frac{25}{2} \cdot 15 = 187.5 \)
- \( R_2 = 3 \cdot 10 = 30 \)
- \( R_3 = \frac{4}{5} \cdot 6 = 4.8 \)

### Final Ratio
Thus, the final ratio of resistances is:

\[ R_1 : R_2 : R_3 = 187.5 : 30 : 4.8 \]